Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double f(double b, double hh, double s) {
  double d = 4 * b * hh * (b * hh - 2 * s);
  d = sqrt(d);
  return (2 * b * hh - d) / (2 * b);
}
int main() {
  long n, h;
  scanf("%ld%ld", &n, &h);
  double s = (double)h / (2 * n);
  double hh = h, b = 1;
  stack<double> st;
  for (long i = 1; i < n; i++) {
    double ans = f(b, hh, s);
    st.push(hh - ans);
    b = (2 * s - b * ans) / ans;
    hh -= ans;
  }
  while (!st.empty()) printf("%.10f ", st.top()), st.pop();
  return 0;
}